Difference between revisions of "2015 AMC 10A Problems/Problem 6"

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<math>\frac{a}{b} = \frac32</math>, so the answer is <math>\boxed{\textbf{(B) }\frac32}</math>.
 
<math>\frac{a}{b} = \frac32</math>, so the answer is <math>\boxed{\textbf{(B) }\frac32}</math>.
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==Solution 2==
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Without loss of generality, let the two numbers be <math>3</math> and <math>2</math>, as this pair clearly satisfies the condition of the problem. The ratio is <math>\boxed{\textbf{(B) }\frac32}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 16:24, 24 April 2022

The following problem is from both the 2015 AMC 12A #4 and 2015 AMC 10A #6, so both problems redirect to this page.

Problem

The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?

$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$

Solution

Let $a$ be the bigger number and $b$ be the smaller.

$a + b = 5(a - b)$.

Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and then solving gives

$\frac{a}{b} = \frac32$, so the answer is $\boxed{\textbf{(B) }\frac32}$.

Solution 2

Without loss of generality, let the two numbers be $3$ and $2$, as this pair clearly satisfies the condition of the problem. The ratio is $\boxed{\textbf{(B) }\frac32}$.

Video Solution

https://youtu.be/JLKoh-Nb0Os

~savannahsolver

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2015 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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